Difference between revisions of "2000 AMC 12 Problems/Problem 20"
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<cmath>(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}</cmath> | <cmath>(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}</cmath> | ||
We subtract <math>(2)</math> from <math>(1)</math> to get that <math>xyz + \frac{1}{xyz} = 2</math>. Hence, by inspection, <math>\boxed{xyz = 1 \rightarrow B}</math>. | We subtract <math>(2)</math> from <math>(1)</math> to get that <math>xyz + \frac{1}{xyz} = 2</math>. Hence, by inspection, <math>\boxed{xyz = 1 \rightarrow B}</math>. | ||
+ | ~AopsUser101 | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 11:01, 6 July 2019
Problem
If and are positive numbers satisfying
Then what is the value of ?
Contents
[hide]Solution
Solution 1
We multiply all given expressions to get: Adding all the given expressions gives that We subtract from to get that . Hence, by inspection, . ~AopsUser101
Solution 2
We have a system of three equations and three variables, so we can apply repeated substitution.
Multiplying out the denominator and simplification yields , so . Substituting leads to , and the product of these three variables is .
Also see
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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